[结题报告]11479 - Is this the easiest problem? Time limit: 1.000 seconds

Problem I
Is this the easiest problem?
Time Limit : 1 second

A triangle is a geometric shape with three positive sides. However, any given three sides won’t necessarily form a triangle. The three sides must form a closed region. Triangles are categorized depending on the values of the sides of a valid triangle. In this problem you are required to determine the type of a triangle.

 

Input

The first line of input will contain a positive integer T<20, where T denotes the number of test cases. Each of the next T lines will contain three 32 bit signed integer.

Output

For each case of input there will be one line of output. It will be formatted as:
Case {x}: {triangle type}. Where x denotes the case number being processed and {triangle type} is the type of the triangle..{triangle type} will be one of the following, depending on the values of the three sides:

Invalid - The three sides can not form a triangle
Equilateral - All three sides of valid triangle are equal
Isosceles - Exactly two of the sides of a valid triangle are equal.
Scalene - No pair of sides are equal in a valid triangle.


Sample Input Sample Output

4
1 2 5
1 1 1
4 4 2
3 4 5

Case 1: Invalid
Case 2: Equilateral
Case 3: Isosceles
Case 4: Scalene

 

参考代码：给定三角形的三条边，判断此三角形的类型（Invalid， Equilateral，Isosceles，Scalene），其符合条件分别为任意2边小于等于第三边；3边全等，任意2边相等，和3边不等。用if语句判断一下就好。

#include"stdio.h"
int main()
{  long n,a,b,c,i=1;
   scanf("%ld",&n);
   while(n--)
   {scanf("%ld%ld%ld",&a,&b,&c);
    if(a+b<=c||a+c<=b||b+c<=a)
    printf("Case %ld: Invalid\n",i);
    else if(a==b&&b==c&&c==a) printf("Case %ld: Equilateral\n",i);
    else if(a==b&&a!=c) printf("Case %ld: Isosceles\n",i);
    else if(a==c&&a!=b) printf("Case %ld: Isosceles\n",i);
    else if(b==c&&b!=a) printf("Case %ld: Isosceles\n",i);
    else if(a!=b&&a!=c&&b!=c) printf("Case %ld: Scalene\n",i);
    i++; 
             }
    }

 

转载于:https://www.cnblogs.com/sjy123/archive/2013/02/21/2920021.html